Problem Solving Strategies

Ice Inkberry

13:00

Is it correct? The angle $\theta$ will be the angle of tangent (with the x axis) at x --> infinity.

harambe

13:24

@sammygerbil are you free?

sammy gerbil

@IceInkberry Yes. The path of the particle is a hyperbola. ie Far from Q in both directions $-\infty$ and $+\infty$ the path is a straight line.

@harambe yes

harambe

@sammygerbil imgur.com/a/1BoMJbx

Q41

What do they mean by resonant frequency? Is it fundamental frequency or just multiples of fundamental frequency

sammy gerbil

@harambe Resonant frequency refers to a standing wave. The lowest frequency would have a standing wave of half a wavelength on the string. I don't understand the significance of the maximum value of the fundamental frequency.

harambe

Okay

The question doesn't specify n though

sammy gerbil

@harambe No. Possibly the question should ask for the minimum value of the fundamental frequency? What do you think?

harambe

13:39

I think that makes sense.. Even I don't see signifacance in maximum fundamental frequency... Isn't fundamental frequency having half wavelength fixed

Abcd

@sammygerbil please ping me when you are free

sammy gerbil

@harambe Yes. That is maximum wavelength / minimum frequency.

harambe

@sammygerbil shouldn't it be reverse

sammy gerbil

@harambe Why?

harambe

Well L=lamda/2 is the shortest wavelength in resonance at fixed ends

@sammygerbil nvm. I got it

You are right

@Abcd I am done

sammy gerbil

13:46

@harambe Perhaps what they mean is that there is resonance (a standing wave) at 240Hz and the next resonance is at 360 Hz.

Abcd

harambe

@sammygerbil I think that too. Gotta try it then.

Abcd

sammy gerbil

@Abcd What is your question / doubt?

Are you asking where you have gone wrong?

You have 2 springs in series so the effective spring constant is given by $$1/k_{eff}=1/k_1+1/k_2$$

Abcd

@sammygerbil that's what I have done

@sammygerbil c 2nd last Line

sammy gerbil

14:01

The tension $F$ in the wire is related to its extension $x$ by $$F=\frac{YA}{L}x=k_1 x$$ so $$\frac{1}{k_{eff}}=\frac{L}{YA}+\frac{1}{k_2}=\frac{kL+YA}{kYA}$$

Abcd

@sammygerbil please tell me my mistake

sammy gerbil

The period is $$T=2\pi\sqrt{\frac{m}{k_{eff}}}=2\pi\sqrt{\frac{(kL+YA)m}{YAk}}$$

@Abcd I think your mistake is that you are using $k$ to mean both the spring constant of the wire and the spring constant of the spring.

Abcd

@sammygerbil No I have called it k2 for the wire

@sammygerbil See the circle with a blue pen

@sammygerbil I have spotted the mistake from your working though

@sammygerbil You are getting $k_{wire} = \dfrac{YA}{l}$

But I am getting it as $\dfrac{lk^2}{AY}$

@sammygerbil Why^?

sammy gerbil

@Abcd I told you. In the first line you have said the spring constant of the wire is $k$. That is wrong, it is the spring constant of the spring.

Abcd

@sammygerbil NO there i mean the force constant of the spring

@sammygerbil Stress is given by F/A

@sammygerbil But F here is kx where k is the spring constant of the spring.

that's why:

$Stress_{wire} =\dfrac{k_{spring}x_{spring} }{A}$

@sammygerbil Spring is exerting $F = kx$ on wire.

sammy gerbil

14:16

@Abcd ok I see.

In that case $x$ relates to the extension of the spring not the wire.

So you cannot say $F=dU/dx$ for the wire, because $x$ in this definition should relate to the wire not the spring.

Abcd

@sammygerbil Oh

@sammygerbil what to do then

sammy gerbil

@Abcd I recommend you to use a simpler solution! Find the force constant for the wire (as I did), then find effective force constant for wire + spring.

Abcd

@sammygerbil okay

sammy gerbil

Actually no calculation is required for this problem. In the limit that the wire becomes very stiff $Y\to \infty$ the period should tend to that of the spring alone.

(b) is the only option which has that limiting value.

Abcd

@sammygerbil Amazing method!! Please keep telling me such tricks whenever they strike to you. JEE needs very quick solving and such tricks are super helpful

sammy gerbil

14:28

ok

The trick is to examine special cases for which you know the answer.

That should enable you to eliminate some options. Pehrhaps all except one, as here.

Abhas Kumar Sinha

15:10

Hello

While Doin' Irodov's Problems in General Physics Chapter - , question number - 13. I found the answer in much simpler and in 3 liner way (without Calculus) $\tau = \frac{l}{v^2 - u^2}$ but the answer given in the Book is = $\frac{vl}{v^2 - u^2}$ also, what I found that answer given in the Book seems dimensionally incorrect. So, what's the true answer?

Dante

@sammygerbil Hello

sammy gerbil

@Dante hello

Abhas Kumar Sinha

@sammygerbil Hello Professor

sammy gerbil

@AbhasKumarSinha Hello. Check again the dimensions / units for both expressions. Your answer does not have the correct dimensions. The book answer does.

Abhas Kumar Sinha

@sammygerbil Sorry Professor, I got the problem here

I made square root twice

Thanks sir

Dante

15:23

A rigid wire is in the shape of a parabola with mass per unit length λ. Equation of parabola is $y=\sqrt{2}x^2$

Find torque about O due to weight of the rod.

Abcd

@Dante \sqrt not /sqrt

Dante

Yeah, got that

Abcd

@Dante $\tau = r\times F$ should be sufficient i guess with tad bit maths

Dante

I have been trying

Didn't get the answer.

Abcd

trying

perpendicular distance is just $x$

$|d\vec \tau| = (dm)xg $

$dm = \lambda dx$

Dante

15:29

I thought the same, but after applying limits I got torque as infinity

Abcd

$d\tau= g x \lambda dx $

Is answer 0 @Dante ?

Dante

Answer must be in the form of $\frac{(2n+1)λg}{2n}$

We gotta find n

Abcd

\dfrac is larger

Dante

@Abcd No Even I don't understand why, it's a question from FIITJEE book

My friend was like we gotta consider only half parabola

Abcd

@Dante which fiitjee book

Dante

15:32

If we consider full torque will be zero.

I have no idea why

@Abcd Gotta ask him

harambe

@sammygerbilp you free for a math question

Abcd

I dont know then

@sammygerbil would be knowing

@sammygerbil are you here??

Dante

@Abcd It's from AIITS

sammy gerbil

@Dante Sorry I don't understand what the problem is asking. If the parabolic wire is symmetrical about O, the resultant torque is zero.

Abcd

@Dante oh i see . yesterday was AI2Ts i i didnt give

@Dante Surely your friend hasnt sent you the full question. Ask him to send a pic

sammy gerbil

15:38

@harambe Not sure if Dante needs more help. But post your problem anyway.

Dante

@Abcd K,

sammy gerbil

@Dante Can you upload an image of the question?

Dante

Yes, on it

sammy gerbil

@Dante Wasn't there a printed diagram? There is space for one.

Dante

Nope, that's for rough work

harambe

15:46

@sammygerbil okay

sammy gerbil

@Dante Strange. I can see a $y$ and $x$ and an arrow. As though something was there which has been erased.

Dante

Oh wait, even I can see that, lemme ask him

But he doesn't have any reason to ease it XD

Abcd

lol

@sammygerbil surely the pic didnt get printed in the exam paper XD

Dante

Yes, you are right

My friend has traced the diagram itself

Abcd

@Dante no Ai2ts never gives such space for rough work

there is rough work space at the end of each page, at the bottom.

Dante

15:51

I didn't know that.

sammy gerbil

@Dante The difficulty for me is that even if we only take the part of the parabola for which x > 0 then there is nothing to prevent it from being infinitely long and therefore infinitely heavy. So the torque would be infinite. I think there must be some information missing, perhaps stated in the diagram which did not get printed.

Dante

I just asked him if there was any limit

Abcd

@Dante just ask him to arrange the pdf of the paper. AI2Ts pdfs are circulated after the exam.

Dante

He says he has traced exactly what was visible.

Abcd

Dont waste time.

Dante

15:52

Right

Abcd

in JEE/High School Chemistry Problems, 2 days ago, by Abcd

1 hour on 1 question is not acceptable at this stage.

sammy gerbil

@Abcd Good advice. It could be a nice problem, but there is something missing and we don't know what it is.

@Dante How did your friend solve it? What assumption(s) did he make?

Dante

He didn't get it

sammy gerbil

@Dante ok. Pity.

Dante

@sammygerbil Infinity is also one of the possibility when n=0 in above problem

So let us just wait until the key and solution is released XD

Abcd

16:02

its released already

ask him the code of the paper

I have all the answer keys

I will post the solution of this question,

ask him the question number too.

@Dante

Dante

k

Abcd

@sammygerbil doubts

sammy gerbil

@Abcd yes?

Dante

Set A

code 116520

Abcd

@sammygerbil A particle of mass m is executing oscillations about the origin on the x axis. Its potential energy is $U(x) = k|x|^3$. where x is a positive constant. If the ampltiude of oscillation is A then its time period T is proportional to:

a) $1/\sqrt a$

b) a^0 (independent of a)

c) $\sqrt a$

d) $a^{3/2}$

Dante

16:16

1st question

sammy gerbil

@Abcd I am trying to find a question on the main site which addressed this issue.

Abcd

sammy gerbil

Maybe we can solve by reasoning. For harmonic oscillator $U=kx^2$ period is independent of amplitude.

Abcd

there is a finite limit.

$x_o$

See the solution above

@sammygerbil option a is eliminated

When $a= 0, T =0$

Not infinite.

So option A gone.

Dante

Hmm, what's the question exactly?

Abcd

16:19

@Dante i just have solution and answer key as I said

Dante

Oh, I see

sammy gerbil

@Abcd That is good thinking. But not necessarily true. ... If we have $U=k|x|^n$ then we might expect period to increase with amplitude for n > 2 and decrease with amplitude for n < 2.

Abcd

@sammygerbil But option A gives infinity man thats plain wrong

@sammygerbil how to solve the problem but

sammy gerbil

@Abcd Period could tend to infinity for small amplitudes if the curve is flat at the origin.

Abcd

@sammygerbil hmm

sammy gerbil

16:27

@Abcd From my argument above I would guess that for n=3 the period decreases as the amplitude increases. Option A!

2

A: Frequency of small oscillation of particle under gravity constrained to move in curve $y=ax^4$

The motion in such a curve is quite hard to calculate, and even more so if you do not want to get into the messy details of Jacobi elliptic functions like $\text{sn}(u|k)$. However, for the case of small oscillations there is a simple scaling argument that lets you calculate the dependence of the...

This is not the question I was thinking of, but it shows that for n=4 the period is proportional to 1/a.

Which agrees with $1/\sqrt{a}$ for n=3.

Abcd

@sammygerbil i dont know what the answer is saying , we hvent been taught all that in maths

@sammygerbil you mean its increasing in powers of $-1/4$

$x -> a^0$

$x^2 -> a^{(-1/4)}$

$x^3 -> a^{(-1/2)}$

Dante

@sammygerbil Can we not write acceleration of particle as $-3kx *\mod{x}$ and find time required to reach equilibrium from extreme using suitable integration process and then multiply it by 4 to get the answer?

Abcd

@Dante why dont you try and tell if you get the answer?

@Dante \times not *

Dante

I tried it, could do it

sammy gerbil

@Abcd The answers say that $T \propto a^{1-n/2}$.

Abcd

16:33

@sammygerbil do you remember a similar problem before and I had done that using dimensional anlsysis?

Dante

couldn't* maybe sammy knows the proper way to integrate

Abcd

@Dante lol what a typo

Dante

Hehe

My friend's asking where you got the key from?

Abcd

1 min ago, by Abcd

@sammygerbil do you remember a similar problem before and I had done that using dimensional anlsysis?

@Dante Its posted in my fiitjee batch group after the exam

@sammygerbil this:

Aug 7 at 4:57, by Abcd

@JohnRennie In exam, I did this question using dimensional analysis. I checked which of the 4 options has dimensions of time and got the right answer ;).

Aug 7 at 15:03, by Abcd

@sammygerbil Wow! Amazing method. Thanks

1

Q: Period $T$ of oscillation with cubic force function

How would I find the period of an oscillator with the following force equation? $$F(x)=-cx^3$$ I've already found the potential energy equation by integrating over distance: $$U(x)={cx^4 \over 4}.$$ Now I have to find a function for the period (in terms of $A$, the amplitude, $m$, and $c$), b...

sammy gerbil

4

Q: Non-SHM oscillatory motion

How to solve these kind of questions , where $|F| \propto x^2$? How to find time period and velocity type related things to the oscillatory motion? $$m\dfrac{d^2x}{dt^2}=F=-\dfrac{dU}{dx}=-3kx|x|.$$ But after this $$m\dfrac{d^2x}{dt^2}=-3kx|x|.$$ What is general solution of this ODE? I thin...

homework-and-exercises newtonian-mechanics forces non-linear-dynamics oscillators

Abcd

16:37

6

A: Finding the period of an anharmonic oscillation by substituting the solution for SHM

Your doubts about the solution given are justified. The method of solution seems to be invalid and misguided - but see my footnote. However, the correct answer choice is still (A). If the potential energy is $V=k|x|^3$ then (as you observe) the motion is not simple harmonic and cannot be descri...

@sammygerbil please tell how to do

@sammygerbil More questions, leave that if you are not getting it right now.

I cant give more than 5 or max 7 minutes to a question.

sammy gerbil

@Abcd There is a solution using dimensional analysis by zkf in the question Non-SHM Oscillatory Motion.

However, I think you can reason as I suggested. As n increases the curve becomes flatter at the origin, so you should expect the period to increase.

Abcd

sammy gerbil

(The curve $y=x^n$ looks more and more like a "square well" for $n\to \infty$.)

Abcd

@sammygerbil zkf solution takes the assumption that time period will be of form $1/\sqrt k$

@sammygerbil anyway,please c the new problem.

Cant give more than 7 minutes to a problem while practicing.

sammy gerbil

16:55

@Abcd And you cannot use the internet if you are practising under exam conditions!

Abcd

@sammygerbil By practicing i meant clearing doubts and solving other questions and stuff. While practicing in exam conditions i do a 3 hour paper without using internet

sammy gerbil

@Abcd I don't understand zkf solution. It might be correct, but it is not clear to me.

Abcd

@sammygerbil same here

@sammygerbil please c next problem now.

sammy gerbil

@Abcd B is incorrect : energy stored is $\frac12 Fx$. C is correct.

D also correct, as well as A of course.

Abcd

Damn it

silly mistake

leave this question

sammy gerbil

17:06

ok

Abcd

sammy gerbil

@Abcd There should be a quick method, but I cannot think what it is.

Abcd

@sammygerbil Okay, we'll take this up later.

sammy gerbil

@Abcd I am not sure I understand the diagram. Is the hemisphere hanging from a support?

Abcd

@sammygerbil Please help with option (T). I cant understand how to decide about (T) for each situation

@sammygerbil No idea ...

sammy gerbil

17:16

@Abcd What is option T? It is cut off at the bottom.

Abcd

@sammygerbil whatever you can see at the bottom. There's nothing after that.

Option T ends at "...length"

@sammygerbil Are you able to read option T?

sammy gerbil

@Abcd Got any ideas?

Abcd

@sammygerbil I was trying energy conservation, but that helps with other options. not option T.

@sammygerbil I also thought of using symmetry of SHM but am not sure of this method

sammy gerbil

@Abcd What answers do you have so far?

Abcd

A- PQRS
B-PQRS
C- T
D- T

@sammygerbil sorrry

C- P

D-P

Not T

Just typed that by mistake in hurry

sammy gerbil

17:33

@Abcd Looks ok. But I'm not getting T. It looks like the 2nd sentence is not complete.

Abcd

@sammygerbil Answers are:

A- PQRST
B- PQRST
C- PQRS
D- PQRS

@sammygerbil so for C and D I think they have considered complete net force as 2mg not 3mg

@sammygerbil but still i dont understand option T for any of A B C and D. Please help

new doubts^

First question of the page and last question.

Attempts:

First one:
dont have much idea.

Second one:

$T' = T/4 + T/8 + T/8 + T/4 = T/2 + T/4 = 3T/4$

sammy gerbil

@Abcd I can understand why T applies to A and B. We are being shown the instant when the force is applied. It starts from rest. But it's not a good question, confusing, not clear.

Abcd

So K = 4/3

But its an integer type question

Answer gotta be an integer

@sammygerbil you mean symmetry of SHM?

@sammygerbil Isnt velocity 0 at the end point of SHM?

Are you considering the point shown as the end point?

sammy gerbil

@Abcd I don't know. Let's come back to this question later.

Abcd

@sammygerbil OK, 2 questions scheduled for later.

sammy gerbil

17:46

@Abcd Spring on block question : eqn of motion is y=Acos(wt). We can find time from y=A at t=0 to y=A/2. Latter occurs at cos(wt)=1/2 so wt=$\pi/3$.

Abcd

@sammygerbil which question?

sammy gerbil

@Abcd "A block is connected to a spring..."

Abcd

@sammygerbil whats wrong with my method

sammy gerbil

The velocity of the block is not constant, so the time taken is not proportional to distance.

The block is fastest near the eqm position and slowest at the extremes.

Going from 0 to A/2 takes less time than from A/2 to A.

Abcd

@sammygerbil Oh damn

generation old silly mistake

@sammygerbil its sin not cos

sammy gerbil

17:56

@Abcd It can be either. Depends at what point you set t=0.

Abcd

So now I am getting $T = \dfrac{4\pi}{3\omega }$

sammy gerbil

I have set t=0 when y=A.

Abcd

I have set it when y = 0

@sammygerbil So now I am getting K = 3/2

But answer is 3

Time to reach A = $\pi/2 \omega$

Time to reach A+ A/2 = $\pi/ 6 \omega$

Time period = $2 \times (\pi/2\omega + \pi/6\omega)$

@sammygerbil are you there

sammy gerbil

@Abcd Continuing my solution, $\omega t=\frac{2\pi}{T}t=\frac{\pi}{3}$. So $t=T/6$. This means the period of oscillation is $T-2T/6=T-T/3=(2/3)T$.

Abcd

@sammygerbil Yes, we both get same answer

@sammygerbil OK, leave it. must be wrong answer in the key.

@sammygerbil please see the first question then

sammy gerbil

18:07

@Abcd The rolling disk?

Abcd

@sammygerbil yes

sammy gerbil

Torque is $Rkx$. Angular acceleration is $I\dot \omega$. And $R\omega=\dot x$.

Abcd

@sammygerbil ??

sammy gerbil

@Abcd What is your difficulty?

Abcd

@sammygerbil torque about ground?

sammy gerbil

18:12

If the spring extends by $x$ for restoring force is $kx$.

@Abcd Yes, the disk rotates about the point of contact with the ground.

Abcd

@sammygerbil there wont be linear shm? why not?

sammy gerbil

Moment of inertia about point of contact with ground is $I=\frac32 MR^2$.

There will be SHM.

Abcd

40 secs ago, by Abcd

@sammygerbil there wont be linear shm? why not?

sammy gerbil

@Abcd You tell me! You are making the assertion.

Abcd

@sammygerbil you are considering only angular shm

sammy gerbil

18:15

@Abcd Angular SHM, linear SHM, makes no difference.

If restoring force is linear with x, motion is SHM.

Abcd

@sammygerbil ok

sammy gerbil

We get : $$\frac32MR^2\dot\omega=\frac32MR\ddot x=-Rkx$$ $$\ddot x+\frac{2k}{3M}x=0$$ So X=2.

@Abcd I am going to take a break for about 30 minutes. Will ping you when I'm back.

Abcd

Ok

sammy gerbil

18:55

@Abcd Back.

@Abcd What answer does the key give if not $t=(2/3)T$?

Abcd

19:25

@sammygerbil 3

sammy gerbil

@Abcd That cannot be correct : it must be greater than $t=\frac12 T$.

Are you satisfied with the question about the rolling disk?

Abcd

@sammygerbil yes

@sammygerbil $y = f(t - x/v)$

@sammygerbil Not getting intuition for this thing^. Why is travelling wave always of this form?

@sammygerbil Book's author has tried to explain but his explanation is super messy and complicated to understand.

sammy gerbil

19:41

@Abcd Short answer : this is the most general solution to the wave equation. Any shape of disturbance which keeps its shape while moving sideways with constant speed, is a solution to the wave equation.

Abcd

@sammygerbil my book says it represents displacement of x - 0 as time passes

sammy gerbil

@Abcd It might help to read through the discussion between JR and Harambe about waves earlier today.

Abcd

@sammygerbil where? link?

sammy gerbil

9 hours ago, by harambe

@JohnRennie http://imgur.com/a/pAjP6eq

:Abcd Starting here ^^^

JD_PM

Hi Abcd and Sammy. @sammygerbil just when you are free, may you ha a look at this? imgur.com/a/6Fa3DMA the 1/2 term comes out of the blue for me; it is not explained why 1/2?

I know it can be obtained experimentally. However, is it possible to get it theoretically?

have*

Abcd

20:03

Hi JD

@sammygerbil Suppose you are travelling with velocity of wave and seeing it. What would you see?

Both wave and you have v velocity.

sammy gerbil

@Abcd You would see a stationary wave. Just as when you travel at the same speed as a car you see a stationary car.

Abcd

@sammygerbil What do you mean by stationary wave

You mean just a graph of sin x?

no movement?

sammy gerbil

@JD_PM $$\frac{d}{dt}\frac12 v.v=\frac12(\frac{dv}{dt}.v+v.\frac{dv}{dt})=\frac12(2v.\frac{dv}{dt})$$

@Abcd Yes, just a graph of sine x, no movement.

Abcd

@sammygerbil suppose you take a small part of it $\Delta L$, would you see circular motion of string particles in $\Delta L$

@sammygerbil please reply

sammy gerbil

20:51

@Abcd If you are travelling along with the wave you do not see any movement of the wave. The particles change, the wave shape does not. Imagine you are watching a Mexican Wave in a football stadium. Imagine you are walking along at the same speed. The peak of the wave is always opposite you, but the person at the peak is constantly changing.

The particles are the medium in which the wave travels. They pass on the wave motion from one to another, like the Mexicans. Each Mexican only goes up and down, the sequence in which they do that is the wave, which travels horizontally.

Nobody recognizeable

23:27

@sammygerbil you there ? I had to ask an differentiation question.

sammy gerbil

@Nobodyrecognizeable yes?

Nobody recognizeable

@sammygerbil any ideas . I used l' hospital to get 0.

@sammygerbil though I gotta apologise as this is a math question.

sammy gerbil

@Nobodyrecognizeable I get 5. It is quite straightforward. Differentiate numerator and denominator, insert the values given.

Nobody recognizeable

@sammygerbil how ?

sammy gerbil

Have another try.

Nobody recognizeable

23:38

@sammygerbil i see f(a) and g(a) are constant. I don't know how that amazed me to take them as variables..

@sammygerbil do you recall leibnitz theorem? If $y= x^(n-1)lnx$

sammy gerbil

@Nobodyrecognizeable No I do not recall.

Nobody recognizeable

@sammygerbil ok. Now some of physics questions.

sammy gerbil

@Nobodyrecognizeable :)

Nobody recognizeable

@sammygerbil as the block rests plank should be giving mg/10 force back.

sammy gerbil

@Nobodyrecognizeable You mean that is the friction force accelerating the block horizontally?

Nobody recognizeable

23:51

@sammygerbil i think so. Another else does not seem to counteract the force. As the block remains at rest the center of mass does not move even though the system is accelerating. Then there is bound to be a second force counteracting.

sammy gerbil

@Nobodyrecognizeable In the accelerating frame of the plank, there is a fictitious inertial force $ma$ acting backwards through the COM of the block. In the inertial frame of the ground there is no need for a counteracting force : there is an unbalanced force on the block which causes acceleration $a$.

Nobody recognizeable

@sammygerbil but the question says the block is at rest wrt the plank.

sammy gerbil

Using the accelerating frame of the plank, there is a horizontal couple (friction, fictitious force) and a vertical couple (weight, normal force) which is equal and opposite, so that the resultant turning moment is zero.

@Nobodyrecognizeable Yes. And the plank is accelerating wrt the ground.

So the block is also accelerating. There must be a force causing that.

Nobody recognizeable

@sammygerbil so ypu wrt to inertial observer's frame both are accelarating in same direction with same magnitude.

sammy gerbil

@Nobodyrecognizeable Yes.

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